trigonometry.pdf

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ADVANCED MATHEMATICS
TRIGONOMETRY
7
BARAKA
LO1BANGUT1
sec2
(X) - tan2
(X) = 1

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BARAKA LO1BANGUT1
5 | b a r a k a l o i b a n g u t i @ g m a i l . c o m
TRIGONOMETRY
The word trigonometric is derived from two Greek
words trigon, which means triangles, and metric,
which means measures. Knowledge of the trigonometrical ratios is vital in very
many fields of engineering, mathematics and physics.
Trigonometry, as the word implies, is concerned with the measurement of the parts,
sides and angles, of a triangle. The early applications of the trigonometric functions
were to surveying, navigation, and engineering. These functions also play an
important role in the study of all sorts of vibratory phenomena-sound, light,
electricity, etc. Therefore, a considerable portion of the subject matter is concerned
with a study of the properties of and relations among the trigonometric functions.
Trigonometric ratios as ratios of a right-angled triangle
c
a
=
=
Hypotenuse
opposite
θ
sin
c
b
=
=
Hypotenuse
Adjacent
θ
cos
cosθ
θ
sin
Adjacent
Opposite
θ
tan =
=
=
b
a
θ
A
C
B
b
c
a
Chapter
7

BARAKA LO1BANGUT1
Reciprocal of trigonometric ratios
Reciprocal of sine is cosecant,
sine
1
cosecant = which is abbreviated as
cosec or csc
Reciprocal of cosine is secant,
cosine
1
secant = , which is abbreviated as
sec.
Reciprocal of tangent is cotangent,
tangent
1
cotangent = , this is
abbreviated as cot
Therefore,









=
=
=
x
x
x
x
x
x
tan
cot
sin
cosec
cos
sec
1
1
1
Trigonometric identities
Consider the triangle ABC given
c
a
θ
sin = and
c
b
θ
cos =
2
2
2






+






=
+
c
b
c
a
θ
cos
θ
sin 2
2
2
2
2
c
b
a
θ
cos
θ
sin
+
=
+
2
By Pythagoras theorem
2
2
2
c
b
a =
+
θ
A
C
B
b
c
a

BARAKA LO1BANGUT1
1
2
=
=
+ 2
2
2
c
c
θ
cos
θ
sin , therefore 1
θ
cos
θ
sin =
+ 2
2
Using 1
θ
cos
θ
sin =
+ 2
2
divide throughout by θ
cos2
θ
cos
θ
cos
θ
sin
θ
cos
θ
cos
2
2
2
2
2
1
=
+ therefore θ
sec
θ
tan 2
=
+ 2
1
Using 1
θ
cos
θ
sin =
+ 2
2
divide throughout by θ
sin 2
θ
sin
θ
sin
θ
sin
θ
sin
θ
cos
2
2 2
2
2
1
=
+ therefore θ
cosec
1
θ
cot 2
=
+
2
SIMPLIFYING TRIGONOMETRIC IDENTITIES
This means to write the compound trigonometric given in a most simplified
form with one or two trigonometric functions.
Simplify



2
cosec
tan
tan 1
2
+
Solution
Given



2
cosec
tan
tan 1
2
+
from θ
sec
θ
tan 2
=
+ 2
1 then
















2
2
1
sin
cos
sin
sec
=
=



sin
cos
sec
1
2
( )



sin
cos
cos

2
1


cos
sin
=
Example 1

BARAKA LO1BANGUT1
Therefore, 



tan
cosec
tan
tan
2
=
+1
2
Simplify x
x
x sin
cos
cosec 2
+
Solution
Given x
x
x sin
cos
cosec 2
+
x
x
x
sin
cos
sin
+





 2
1
x
x
x
x
x
cosec
sin
sin
sin
cos
=
=
+ 1
2
2
Therefore, x
x
x
x cosec
sin
cos
cosec 2
=
+
Simplify
1
1
1
1
+
+
− x
x sec
sec
Solution
Given
1
1
1
1
+
+
− x
x sec
sec
( ) ( )
( )( )
1
1
1
1
+
−
−
+
+
=
x
x
x
x
sec
sec
sec
sec
1
2
2
−
=
x
x
sec
sec
x
x
2
2
tan
sec
=
x
x
x
x
x
x
x
cot
cosec
sin
cos
sin
cos
cos
2
2
2
2
2
2
=
=














=
Therefore, x
x
x
x
cot
cosec
sec
sec
2
1
1
1
1
=
+
+
−
Example 2
Example 3

BARAKA LO1BANGUT1
EXERCISE 1
(a) Simplify the following trigonometric identities
1.
x
x
x
csc
sin
tan +
2.
θ
θ
θ
2
cot
sin
csc −
3. ( )
x
x
x
x sin
tan
cos
cot + 4. 

 tan
sin
cos +
5. x
x
x 2
2
2
sin
cos
sin − 6.
x
x
x
x
sin
cos
cos
sin
+
+
+
1
1
7.
x
x
x
tan
csc
sec
+
+
1
8. ( )( )
x
x
x tan
sec
sin +
−
1
9. ( ) ( )2
2
x
x
x
x cos
sin
cos
sin −
+
+
10. ( )( )
1
+
− x
x
x csc
tan
sec 11. ( ) ( )2
2
1
1 x
x tan
tan −
+
+
12. ( ) ( )2
2
3
3 x
x
x
x sin
cos
sin
cos −
+
+
13. ( ) ( )2
2
3
4
4
3 x
x
x
x sin
cos
sin
cos −
+
+
14. x
x 2
2
sin
tan − 15.
x
x
x
2
2
2
tan
cot
csc −
16.
x
x
x
2
1 cos
cos
sin
−
17. x
x
x cos
sin
tan −
18.
x
x 2
2
1
1
tan
sin
−
19. ( )
x
x
x sin
csc
sin −
20.
x
x
x
x
csc
sin
sec
cos
+
(b) From each of equation set given eliminate 
21. 
sin
3
=
x and 
cosec
=
y 22. 
sin
+
= 3
x and 
tan
=
y
23. 
sin
=
x
5 and 
cos
2
=
y 24. 
sin
2
=
x and 
cos
=
y
3
25. 
sin
+
= 3
x and 
cosec
=
y 26. θ
x cot
= and 
sec
+
= a
y

BARAKA LO1BANGUT1
PROOF OF TRIGONOMETRIC IDENTITIES
The first part above was to simplify trigonometric identities, in this part, we
are required to verify that the right-hand side is equals to the left hand side.
We are required to simplify one side to be the same as the other side. To
verify, we can start with either side.
Show that x
x
x
x cos
tan
sin
sec =
−
Solution
x
x
x
x cos
tan
sin
sec =
−
Consider left hand side






−
x
x
x
x cos
sin
sin
cos
1
x
x
cos
sin 2
1−
= x
x
x
cos
cos
cos
=
=
2
, hence LHs become as
RHs, hence shown.
Prove that x
x
x
x cosec
sec
cot
tan =
+
Solution
x
x cot
tan +
x
x
x
x
sin
cos
cos
sin
+ x
x
x
x
x
x
x
x
sec
csc
cos
sin
cos
sin
cos
sin
=
=
+
=
1
2
2
hence shown.
Prove that
x
x
x
x
sin
cos
cos
sin −
=
+
1
1
Solution
Example 4
Example 5
Example 6

BARAKA LO1BANGUT1
x
x
x
x
cos
cos
cos
sin
−
−

+ 1
1
1
(Multiplying by x
cos
−
1 in numerator and
denominator)
( ) ( )
x
x
x
x
x
x
2
2
1
1
1
sin
cos
sin
cos
cos
sin −
=
−
−
x
x
sin
cos
−
=
1
hence shown
Prove that
x
x
x
x
x
cos
sin
cos
tan
cot
1
2 2
−
=
−
Solution
Consider the left hand side
x
x tan
cot −
x
x
x
x
x
x
x
x
cos
sin
sin
cos
cos
sin
sin
cos 2
2
−
=
−
( )
x
x
x
x
x
x
x
cos
sin
cos
cos
sin
cos
cos 1
2
1 2
2
2
−
=
−
−
EXERCISE 2
Prove each of the following identities
1. x
x
x
x csc
tan
cos
csc =
+
2
2. 



tan
cos
tan
sin
=
+
+
1
3.








cos
sin
sin
cos
sin
cos
tan
cot −
=
+
−
4.




tan
sec
cos
sin
+
=
− 1
1
5. 



 2
2
1 cos
cot
tan
cot
tan
−
=
+
−
6. ( )2
1
1
x
x
x
x
tan
sec
csc
csc
+
=
−
+
7. ( )( ) ( )2
1
1 x
x
x
x tan
cos
sin
csc +
=
+
− 8. x
x
x
x 2
2
2
2
csc
sec
cot
sin =
+
+
9. ( ) 1
1 2
2
2
−
=
+ x
x
x sec
tan
sin 10. 1
2 2
2
2
−
=
− x
x
x cos
sin
cos
Example 7

BARAKA LO1BANGUT1
11.
x
x
x
x 2
2
2
2
sin
sec
csc
sec =
+
12. x
x
x
x tan
sin
cos
sec =
−
13. 


 cot
cos
sin
csc =
− 14. x
x
x
x 2
2
2
2
sin
tan
sin
tan =
−
15. 


 2
2
sec
cot
tan
tan =
+
16. x
x
x
x
sec
tan
sin
sin
2
1
1
1
1
=
+
−
−
17.
1
1
1
1
+
−
=
+
−




cot
cot
tan
tan
18.
1
1
+
=
−
x
x
x
x
sec
tan
tan
sec
19. x
x
x
x
cot
csc
cos
cos
−
=
+
−
1
1
20. 



sin
tan
sec
cos
−
=
+
1
21.




sec
sec
sin
sec
+
=
−
1
1 2
2
22. ( ) ( ) 2
2
2
2
b
a
a
b
b
a +
=
+
+
+ 


 sin
cos
sin
cos
23.
x
x
x
x
x
sec
cos
sin
cos
sin
−
=
−
+
−
+
1
1
3
3
1
2
2
2
24. If 
sin
+
= 2
x and 
cos
=
+1
y show that 0
4
2
4
2
2
=
+
+
−
+ y
x
y
x
TRIGONOMETRIC EQUATION
Trigonometric equations may have infinity solutions; to be specific the range
of the solution should be given. The range may be 


 360
0  ,




− 360
360  or 



− 180
180  etc. if not given we give out the
general solution. Note that, this range may also be given in radians. We shall
also refer the four quadrants of the xy-plane to get the required principal
solutions.
Solve for x from 


 360
0 x ; ( ) 3
0
15 .
sin −
=

+
x
Solution
Example 8

BARAKA LO1BANGUT1
Given ( ) 3
0
15 .
sin −
=

+
x
( )
3
.
0
sin
15 1
−
=

+
x

−
=

+ 46
17
15 .
x


=

+ 342.54
,
.46
197
15
x


= 327.54
,
.46
182
x
Solve ( ) 5
0
10 .
sin −
=

+
x from 


 360
0 x
Solution
( ) 5
0
10 .
sin −
=

+
x
( )
5
0
10 1
.
sin−
−
=

+
x

−
=

+ 30
10
x then 

=

+ 330
210
10 ,
x
Therefore 

= 320
,
200
x
Solve 4
0.
cos −
=
x from 



− 180
180 x
Solution
4
0.
cos −
=
x
( )
4
0
1
.
cos −
= −
x

= 58
113.
x , or 
−
= 58
113.
x
Solve 0
1
3 =
+

tan from 



− 180
180 
Solution
Example 9
Example 10
Example 11

BARAKA LO1BANGUT1
Given 0
1
3 =
+

tan
3
1
−
=

tan






−
= −
3
1
1
tan


−
= 43
18.
 or 
= 57
161.

Solve 0
2
=
+ x
x
x cos
sin
sin from 


 360
0 x
Solution
0
2
=
+ x
x
x cos
sin
sin
( ) 0
=
+ x
x
x cos
sin
sin
Case 1: 0
=
x
sin
( )
0
sin 1
−
=
x

= 0
x , 
180 or 
360
Case 2: 0
=
+ x
x cos
sin
x
x cos
sin −
= therefore 1
−
=
x
tan
( )
1
1
−
= −
tan
x

=135
x or 
315
Therefore, 
= 0
x , 
135 , 
180 , 
315 or 
360
Solve 0
1
6 2
=
−
− x
x sin
cos from 


 360
0 x
Solution
Example 12
Example 13

BARAKA LO1BANGUT1
0
1
6 2
=
−
− x
x sin
cos
( ) 0
1
sin
sin
1
6 2
=
−
−
− x
x
0
1
sin
sin
6
6 2
=
−
−
− x
x
0
5
sin
sin
6 2
=
+
−
− x
x
( )( ) 0
1
sin
5
sin
6 =
+
− x
x
Case 1: 0
5
sin
6 =
−
x
6
5
sin =
x  





= −
6
5
sin 1
x

= 44
.
56
x or 
56
.
123
Case 2: 1
sin −
=
x
( )
1
sin 1
−
= −
x

= 270
x
Therefore, 
= 44
.
56
x , 
56
.
123 or 
270
Solve x
x
x cos
3
cos
sin
4 = for 


 360
0 x
Solution
x
x
x cos
3
cos
sin
4 =
0
cos
3
cos
sin
4 =
− x
x
x
( ) 0
3
cos
4
sin =
−
x
x
Case 1: 0
sin =
x
Example 14

BARAKA LO1BANGUT1
( )
0
sin 1
−
=
x

= 0
x , 
180 or 
360
Case 2: 0
3
cos
4 =
−
x
4
3
cos =
x






= −
4
3
cos 1
x

= 59
.
48
x or 
41
.
131
The solutions are 
= 0
x , 
59
.
48 , 
41
.
131 , 
180 or 
360
Solve 1
tan
1
tan
2 =
−

 from 


 360
0 
Solution

 tan
1
tan
2 2
=
−
0
1
tan
tan
2 2
=
−
− 

( )( ) 0
1
tan
2
1
tan =
+
− 

Case 1: 1
tan =

( )
1
tan 1
−
=

Example 15
1
tan
1
tan
2 =
−



BARAKA LO1BANGUT1

= 45
 or 
225
Case 2:
2
1
tan −
=







−
= −
2
1
tan 1


= 44
.
153
 or 
56
.
206
The solution are 
= 45
 
44
.
153 , 
225 , 
56
.
206 or 
225
Solve 

 cos
sin
2
sin
3 2
= from 


 360
0 
Solution


 cos
sin
2
sin
3 2
=
0
cos
sin
2
sin
3 2
=
− 


( ) 0
cos
2
sin
3
sin =
− 


Case 1: 0
sin =


= 0
 , 
180 or 
360
Case 2: 
 cos
2
sin
3 =
3
2
tan =







= −
3
2
tan 1


= 69
.
33
 or 
31
.
146
The solutions are 
= 0
 , 
69
.
33 , 
31
.
146 , 
180 or 
360
Example 16

BARAKA LO1BANGUT1
EXERCISE 3
(a) Solve the following from 
0 to 
360 inclusive
1. 7
.
0
2
cos =
x
2. 0
1
3
tan
2 =
−
x
3. 0
cos
sin
cos2
=
− 


4. 0
1
sin
2
sin
5 2
=
−
− 

5. 0
cos
3
sin
2 =
− 

6. 0
3
cos
5 =
+

7. ( ) 0
cos
sin
2
cos =
+ 


8. x
x
x sin
3
cos
sin
4 =
9. ( )( ) 0
1
sin
1
sin
2 =
+
− 

10. 0
cos
sin
2
sin =
− 


11. ( ) 2
30
tan =

+
x
12. 

 csc
cos
cos
4 2
=
13. 7
sec
3
tan 2
2
=
+ 

14. 0
2
sin
5
sin
2 2
=
+
− x
x
15. x
x 2
tan
8
1
tan
2 =
+
(b) Solve the following from 
−180 to 
180 inclusive
16. x
x 3
cos
2
sin =
17. 0
1
sin
4
sin
2 2
=
+
− x
x
18. x
x
x cos
sin
2
sin
3 2
=
19. ( )
2
1
30
cos =

−
x
20. x
x
x tan
2
sec
3
cos
4 =
−
21. 0
1
cos
sin2
=
+
+ x
x
22. x
x 2
cos
4
sin
4 =
−
23. x
x 2
sin
3
cos
5
5 =
−
24. 3
cos
7
cos2
=
+ x
x
25. x
x
x sin
cos
3
tan
8 =
COMPOUND ANGLES

BARAKA LO1BANGUT1
COMPOUND ANGLES, NATURAL SINE ( )
B
A
sin 
Before we proceed with the compound angles, we need first to recall the
sine rule
Consider the triangle below
Area of triangle ABC, bh
A
2
1
=
c
h
Hyp
opp
sin =
=
A A
c
h sin
=

A
bc
A sin
2
1
=
Using this area formula, we can deduce the sine rule as follows
B
ac
A
bc
C
ab
A sin
2
1
sin
2
1
sin
2
1
=
=
=
Simplifying
B
ac
A
bc
C
ab sin
2
1
sin
2
1
sin
2
1
=
=
Divide throughout by abc,
B
ac
A
bc
C
ab sin
sin
sin =
=
abc
B
ac
abc
A
bc
abc
C
ab sin
sin
sin
=
=
This is sine rule
Sine rule is used to solve triangle problem of the two cases below
(a) Solving ASA (Angle, Side, Angle) and AAS (Angle, Angle, Side) Cases
(b) Solving the Ambiguous SSA (Side, Side, Angle) Case
b
B
a
A
c
C sin
sin
sin
=
=
A
B
C
h

BARAKA LO1BANGUT1
Using the above sine rule
Consider again the triangle ABC below
Area of triangle ABC = Area of triangle ADB + Area of triangle DBC
Area of ABC, ( )
B
A
ac
A +
= sin
2
1
Area of triangle ADB, A
hc
A sin
2
1
= but A
c
h
c
h
A cos
cos =

=
Area of triangle DBC, B
ah
A sin
2
1
= but B
a
h
a
h
B cos
cos =

=
Area of triangle ABC = Area of triangle ADB + Area of triangle DBC
( ) B
ah
A
hc
B
A
ac
A sin
2
1
sin
2
1
sin
2
1
+
=
+
=
( ) B
ah
A
hc
B
A
ac sin
sin
sin +
=
+
( ) ( )( ) ( )( )
A
c
B
a
A
c
B
a
B
A
ac cos
sin
sin
cos
sin +
=
+
( ) B
A
ac
B
A
ac
B
A
ac sin
cos
cos
sin
sin +
=
+
( ) B
A
B
A
B
A sin
cos
cos
sin
sin +
=
+
Therefore,
A
B
C
h
D
A B
c
a
b
( ) B
A
B
A
B
A sin
cos
cos
sin
sin +
=
+

BARAKA LO1BANGUT1
Sine and Tangent are odd functions, thus ( ) A
A sin
sin −
=
− and
( ) A
A tan
tan −
=
− while cosine is an even function, thus
( ) A
A cos
cos =
− , this apply to secant, cosecant and cotangent.
Using the concept of odd function of sine, we can proof ( )
B
A
sin −
Replace, B with B
− ,
( )
( ) ( ) ( )
B
A
B
A
B
A −
+
−
=
−
+ sin
cos
cos
sin
sin
( ) B
A
B
A
B
A sin
cos
cos
sin
sin −
=
−
Therefore,
COMPOUND ANGLES, NATURAL COSINE ( )
B
A
cos 
Before showing these compound angles, we shall first derive the cosine
formula, which will be our base in proving the cosine compound angles
Consider the triangle ABC below
Using Pythagoras theorem, in triangle DBC
( ) 2
2
2
a
h
x
b =
+
−
2
2
2
2
2 a
h
x
bx
b =
+
+
−
( ) B
A
B
A
B
A sin
cos
cos
sin
sin −
=
−
A C
h
D
B
c a
b – x
x

BARAKA LO1BANGUT1
A
c
x
c
x
A cos
cos =

=
A
c
h
c
h
A sin
sin =

=
( ) ( ) ( ) 2
2
2
2
sin
cos
cos
2 a
A
c
A
c
A
c
b
b =
+
+
−
2
2
2
2
2
2
sin
cos
cos
2 a
A
c
A
c
A
bc
b =
+
+
−
( ) 2
2
2
2
2
sin
cos
cos
2 a
A
A
c
A
bc
b =
+
+
−
A
bc
c
b
a cos
2
2
2
2
−
+
=
This is cosine rule,
Note that, cosine rule is used in two cases
(a) Solving the SAS (Side, Angle, Side) Case
(b) Solving the SSS (Side, Side, Side) Case
Now, consider the triangle ABC, let b – x be y
( ) ( )
B
A
ab
b
a
y
x +
−
+
=
+ cos
2
2
2
2
( )
B
A
ab
b
a
y
xy
x +
−
+
=
+
+ cos
2
2 2
2
2
2
In triangle ADC: 2
2
2
y
b
h −
=
(Pythagoras theorem)
In triangle DBC: 2
2
2
x
a
h −
=
( ) ( ) ( )
B
A
ab
xy
b
y
a
x +
−
=
−
−
−
+
−
− cos
2
2
2
2
2
2
( )
B
A
ab
xy
h
h +
−
=
+
−
− cos
2
2
2
2
A
bc
c
b
a cos
2
2
2
2
−
+
=

BARAKA LO1BANGUT1
( )
B
A
ab
xy
h +
−
=
+
− cos
2
2
2 2
( )
B
A
ab
xy
h +
−
=
+
− cos
2
( ) 2
cos h
xy
B
A
ab −
=
+
( )
ab
h
ab
xy
B
A
2
cos −
=
+
( ) 











−












=
+
b
h
a
h
b
y
a
x
B
A
cos
But
b
y
A =
cos ,
a
x
B =
cos ,
b
h
A =
sin and
a
h
B =
sin
( ) B
A
B
A
B
A sin
sin
cos
cos
cos −
=
+
Therefore,
Using the concept of odd and even functions, we deduce the formula
By replacing B with B
− ,
( ) B
A
B
A
B
A sin
sin
cos
cos
cos +
=
−
From the compound angles ( )
B
A
sin and ( )
B
A
cos we deduce the
compound angles of ( )
B
A
tan
( ) ( )
( )
B
A
B
A
B
A
+
+
=
+
cos
sin
tan
( ) B
A
B
A
B
A sin
sin
cos
cos
cos +
=
−
( ) B
A
B
A
B
A sin
sin
cos
cos
cos −
=
+

BARAKA LO1BANGUT1
( )
B
A
B
A
B
A
B
A
B
A
sin
sin
cos
cos
sin
cos
cos
sin
tan
−
+
=
+
( )
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
cos
cos
sin
sin
cos
cos
cos
cos
cos
cos
sin
cos
cos
cos
cos
sin
tan
−
+
=
+
( )












−






+






=
+
B
B
A
A
B
B
A
A
B
A
cos
sin
cos
sin
1
cos
sin
cos
sin
tan
( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
−
+
=
+
Therefore,
For ( )
B
A−
tan
( ) ( )
( )
B
A
B
A
B
A
−
−
=
−
cos
sin
tan
( )
B
A
B
A
B
A
B
A
B
A
sin
sin
cos
cos
sin
cos
cos
sin
tan
+
−
=
−
( )
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
cos
cos
sin
sin
cos
cos
cos
cos
cos
cos
sin
cos
cos
cos
cos
sin
tan
+
−
=
+
( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
−
+
=
+

BARAKA LO1BANGUT1
( )












+






−






=
−
B
B
A
A
B
B
A
A
B
A
cos
sin
cos
sin
1
cos
sin
cos
sin
tan
( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
+
−
=
−
Therefore,
PROVING TRIGONOMETRIC IDENTITIES
Show that ( ) ( ) B
A
B
A
B
A 2
2
cos
cos
cos
cos −
=
−
+
Solution
( ) ( ) B
A
B
A
B
A 2
2
cos
cos
cos
cos −
=
−
+
( )( )
B
A
B
A
B
A
B
A sin
sin
cos
cos
sin
sin
cos
cos +
−
B
A
B
A 2
2
2
2
sin
sin
cos
cos −
( ) ( ) B
A
B
A 2
2
2
2
sin
cos
1
sin
1
cos −
−
−
B
A
B
B
A
A 2
2
2
2
2
2
sin
cos
sin
sin
cos
cos +
−
−
B
A 2
2
sin
cos − , hence proved
( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
+
−
=
−
Example 17
Example 18

BARAKA LO1BANGUT1
Show that ( ) A
A cos
90
sin =
−

Solution
( ) A
A cos
90
sin =
−

( ) A
A
A sin
90
cos
cos
90
sin
90
sin 
−

=
−

But 1
90
sin =
 and 0
90
cos =

( ) ( ) ( ) A
A
A sin
0
cos
1
90
sin −
=
−

( ) A
A cos
90
sin =
−
 , Hence shown.
Show that
( )
( ) B
A
B
A
B
A
B
A
tan
tan
tan
tan
sin
sin
−
+
=
−
+
Solution
Consider the left hand side
( )
( ) B
A
B
A
B
A
B
A
B
A
B
A
sin
cos
cos
sin
sin
cos
cos
sin
sin
sin
−
+
=
−
+
B
A
B
A
B
A
B
A
B
A
B
A
B
A
B
A
cos
cos
sin
cos
cos
cos
cos
sin
cos
cos
sin
cos
cos
cos
cos
sin
−
+
=
B
B
A
A
B
B
A
A
cos
sin
cos
sin
cos
sin
cos
sin
−
+
=
Example 19

BARAKA LO1BANGUT1
B
A
B
A
tan
tan
tan
tan
−
+
= Hence shown
EXERCISE 4
(a) Prove the following identities
1. ( ) ( ) B
A
B
A
B
A cos
sin
2
sin
sin =
−
+
+
2.
( )
( ) B
A
B
A
B
A
B
A
tan
tan
1
tan
tan
1
cos
cos
−
+
=
−
+
3.
( ) ( ) ( )
C
A
C
A
C
B
C
B
B
A
B
A
cos
cos
sin
cos
cos
sin
cos
cos
sin −
=
−
+
−
4. ( ) A
A cos
π
cos −
=
−
5. A
A cos
2
π
sin −
=






+
6. If C
B
A =
+ , B
p
A tan
tan = , show that ( ) C
p
p
B
A sin
1
1
sin
+
−
=
−
7. If a
y
x =
+ sin
sin and b
y
x =
+ cos
cos show that
( ) 1
2
2
cos
2
2
−
+
=
−
b
a
y
x
8. The roots of the equation 0
15
26
8 2
=
+
− x
x are A
tan and B
tan find
(a) ( )
B
A+
cos and (b) ( )
B
A−
sin .
(b) Prove the following
9.
( )
( )
A
B
B
A
B
B
A
tan
tan
tan
1
tan
tan
=
+
+
−
+
10.
( )
( ) B
A
B
A
B
A
B
A
cot
cot
1
cot
cot
cos
sin
+
+
=
−
+
11.
( )
( ) B
A
B
A
B
A
B
A
tan
tan
tan
tan
sin
sin
−
+
=
−
+

BARAKA LO1BANGUT1
SIMPLIFYING TRIGONOMETRIC IDENTITIES
Simplify
( )
y
x
y
x
cos
cos
sin −
Solution
( )
y
x
y
x
cos
cos
sin −
y
x
y
x
y
x
cos
cos
sin
cos
cos
sin −
=
y
x
y
x
y
x
y
x
cos
cos
sin
cos
cos
cos
cos
sin
−
=
( ) y
x
y
x
y
x
tan
tan
cos
cos
sin
−
=
−
Simplify
( ) ( )
B
B
A
B
B
A
cos
cos
sin
sin −
+
−
Solution
( ) ( )
B
B
A
B
B
A
cos
cos
sin
sin −
+
−
B
B
A
B
A
B
B
A
B
A
cos
sin
sin
cos
cos
sin
sin
cos
cos
sin +
+
−
=
B
B
A
A
A
B
B
A
cos
sin
sin
cos
cos
sin
cos
sin
+
+
−
=
Example 20
Example 21

BARAKA LO1BANGUT1
B
B
A
B
B
A
cos
sin
sin
sin
cos
sin
+
=
B
B
B
A
B
A
cos
sin
sin
sin
cos
sin 2
2
+
=
( )
B
B
B
B
A
cos
sin
sin
cos
sin 2
2
+
=
B
B
A
sin
cos
sin
=
Simplify


−

+

15
tan
30
tan
1
15
tan
30
tan
Solution


−

+

15
tan
30
tan
1
15
tan
30
tan
Recall that ( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
−
+
=
+
( )

+

=


−

+

15
30
tan
15
tan
30
tan
1
15
tan
30
tan

=


−

+

45
tan
15
tan
30
tan
1
15
tan
30
tan
1
15
tan
30
tan
1
15
tan
30
tan
=


−

+

Example 22

BARAKA LO1BANGUT1
Simplify ( ) ( )
x
y
y
x −
−
− tan
tan
Solution
Tangent is an odd function, this mean that ( ) ( )
A
A tan
tan −
=
−
( ) ( )
x
y
y
x −
−
− tan
tan
( ) ( )
( )
y
x
y
x −
−
−
−
= tan
tan
( ) ( )
y
x
y
x −
+
−
= tan
tan
( )
y
x−
= tan
2
y
x
y
x
tan
tan
1
tan
tan
2
+
−
=
Deriving double angles from compound angles
From ( ) B
A
B
A
B
A sin
cos
cos
sin
sin +
=
+
Let A
B =
( ) A
A
A
A
A
A sin
cos
cos
sin
sin +
=
+
Therefore, A
A
A cos
sin
2
2
sin =
From ( ) B
A
B
A
B
A sin
sin
cos
cos
cos −
=
+
Let A
B =
( ) A
A
A
A
A
A sin
sin
cos
cos
cos −
=
+
DOUBLE ANGLES FORMULAE
Example 23

BARAKA LO1BANGUT1
Therefore, A
A
A 2
2
sin
cos
2
cos −
=
From ( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
−
+
=
+
( )
A
A
A
A
A
A
tan
tan
1
tan
tan
tan
−
+
=
+
A
A
A 2
tan
1
tan
2
2
tan
−
=
From the double angle formulae, we deduce the half angles.
From A
A
A cos
sin
2
2
sin = replace A by
2
A
The half angle formula for sine, 











=
2
cos
2
sin
2
sin
A
A
A
From A
A
A 2
2
sin
cos
2
cos −
=
Half angle for cosine, 





−






=
2
sin
2
cos
cos 2
2 A
A
A
From
A
A
A 2
tan
1
tan
2
2
tan
−
= replace A by
2
A
Half angle formula for tangent






−






=
2
tan
1
2
tan
2
tan
2 A
A
A
HALF ANGLE FORMULA

BARAKA LO1BANGUT1
Prove that A
A
A
2
cos
tan
1
tan
1
2
2
=
+
−
Solution
A
A
A
2
cos
tan
1
tan
1
2
2
=
+
−
( )
( ) A
A
A
A
A
A
A
A
A
A
2
2
2
2
2
2
2
2
2
2
cos
sin
cos
cos
sin
cos
cos
sin
1
cos
sin
1
+
−
=
+
−
=
A
A
A
A
2
2
2
2
sin
cos
sin
cos
+
−
=
A
A
A 2
cos
sin
cos 2
2
=
−
= , hence shown
Show that 

 2
tan
2
cos
1
2
cos
1
=
+
−
Solution


 2
tan
2
cos
1
2
cos
1
=
+
−



2
2
2
2
sin
cos
1
sin
cos
1
−
+
+
−
=
Example 24
Example 25

BARAKA LO1BANGUT1



2
2
2
cos
1
cos
1
sin
2
+
−
+
=


 2
2
2
tan
cos
2
sin
2
=
= shown
Prove that 

 cot
2
cot
2
csc =
+
Solution


 cot
2
cot
2
csc =
+

 2
tan
1
2
sin
1
+
=


 2
sin
2
cos
2
sin
1
+
=


2
sin
2
cos
1+
=




cos
sin
2
sin
cos
1 2
2
−
+
=
( )




sin
cos
2
cos
1
cos
1 2
2
−
−
+
=






cot
sin
cos
cos
sin
2
cos
2 2
=
=
= Shown
Example 26

BARAKA LO1BANGUT1
Prove that 


tan
2
cos
1
2
sin
=
+
Solution



tan
2
cos
1
2
sin
=
+




2
2
sin
cos
1
cos
sin
2
−
+
= thus
( )




2
2
cos
1
cos
1
cos
sin
2
−
−
+
=



2
cos
2
cos
sin
2
=



tan
cos
sin
=
= , Shown.
Show that x
x
x 2
2
sin
3
2
2
cos
cos −
=
+
Solution
x
x
x 2
2
sin
3
2
2
cos
cos −
=
+
x
x 2
cos
cos2
+
x
x
x 2
2
2
sin
cos
cos −
+
=
x
x
x 2
2
2
sin
sin
1
sin
1 −
−
+
−
=
x
2
sin
3
2 −
= , Hence shown
Example 27
Example 28

BARAKA LO1BANGUT1
EXERCISE 5
Prove the following identities
1.
A
A
A
cos
1
sin
2
tan
+
=






2.






+






=
2
tan
1
2
tan
2
sin
2 


3. 

 tan
cot
2
csc
2 =
−
4. 


 2
sin
sin
2
tan
2
2
tan 2
=
−
5. 




tan
2
cos
cos
1
2
sin
sin
=
+
+
+
6. 

 4
4
sin
cos
2
cos −
=
7. 

 tan
2
cot
2
cot =
−
8. x
x
x
x
x
sin
cos
sin
cos
2
cos
+
=
−
9.
( ) ( )
y
x
y
y
x
y
y
x
2
sin
sin
2
cos
cos
sin
sin
=
−
+
−
10.
( ) y
x
y
x
y
x
cot
tan
cos
cos
cos
+
=
−
11. 

 3
sin
4
sin
3
3
sin −
=
12. 

 3
cos
4
cos
3
3
cos =
+
13. ( ) 3
cot
sin
csc
sin 2
2
2
+
+
=
+ A
A
A
A
14. ( )



 2
2
sin
cos
3
sin
3
sin −
=
15. 

 2
cot
2
tan
cot =
−
16. ( )
B
A
B
A
B
A
tan
tan
1
sec
sec
sec
−
=
+

BARAKA LO1BANGUT1
17. If 
=
+ 45
2 B
A prove that
A
A
A
A
B
tan
2
tan
1
tan
2
tan
1
tan 2
2
+
−
−
−
=
18. If ( ) ( )
y
x
y
x −
=
+ cos
2
sin show that
y
y
x
tan
2
1
tan
2
tan
−
−
=
19.



sin
cos
1
2
tan
−
=






20. If
12
5
tan =
 find (a) 
2
tan (b) 
3
tan
21. If 
=
+ 45
B
A show that
B
B
B
B
A
sin
cos
sin
cos
tan
+
−
=
SIMPLIFYING TRIGONOMETRIC IDENTITIES INVOLVING
DOUBLE ANGLES
Simplify
A
A
2
sin
1
2
sin
1
−
+
Solution
A
A
2
sin
1
2
sin
1
−
+
A
A
A
A
2
sin
1
2
sin
1
2
sin
1
2
sin
1
+
+

−
+
=
( )
A
A
2
sin
1
2
sin
1
2
2
−
+
=
( )
A
A
A
A
2
cos
2
sin
1
2
cos
2
sin
1
2
2
+
=
+
=
Example 29

BARAKA LO1BANGUT1
A
A
A
A
2
2
sin
cos
cos
sin
2
1
−
+
=
A
A
A
A
A
A
2
2
2
2
sin
cos
cos
sin
2
sin
cos
−
+
+
=
( ) ( )( )
( )( )
A
A
A
A
A
A
A
A
A
A
A
A
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
sin
cos
2
2
2
−
+
+
+
=
−
+
=
A
A
A
A
A
A
tan
1
tan
1
sin
cos
sin
cos
−
+
=
−
+
=
Simplify
x
x
sin
2
sin
Solution
x
x
x
x
cos
2
sin
cos
sin
2
=
Simplify ( ) ( ) y
y
x
y
y
x sin
cos
cos
sin −
+
−
Solution
( ) ( ) y
y
x
y
y
x sin
cos
cos
sin −
+
−
Let A
y
x =
−
Example 30
Example 31
x
x
sin
2
sin

BARAKA LO1BANGUT1
y
A
y
A sin
cos
cos
sin +
=
( )
y
A+
= sin but A
y
x =
−
( )
y
y
x +
−
= sin
x
sin
=
EXERCISE 6
Simplify the following questions
1.
A
A
cos
2
sin
2. A
A 2
cos
tan
2
3. A
A 2
cos
sin
2 2
+
4. A
A sin
4
2
cos
3 +
5.
x
x
x
x
sin
cos
1
cos
1
sin +
+
+
6.
x
x
x
x
2
cos
cos
3
1
2
sin
sin
3
+
+
+
7. A
A
A cos
6
π
sin
6
π
sin =






−
−






+
8.
2
cos
1 A
−
9.
A
A
cos
1
cos
1
−
+
10. A
A 2
tan
cos2
+

BARAKA LO1BANGUT1
SOLVING EQUATION INVOLVING DOUBLE ANGLE
Solve 0
cos
2
sin
3 =
+ x
x from 


 360
0 x
Solution
0
cos
2
sin
3 =
+ x
x
0
cos
cos
sin
6 =
+ x
x
x
( ) 0
1
sin
6
cos =
+
x
x
Case 1: 0
cos =
x
( )
0
cos 1
−
=
x

=90
x or 
270
Case 2: 0
1
sin
6 =
+
x
6
1
sin −
=
x






= −
6
1
sin 1
x

= 59
.
189
x or 
41
.
350
Therefore, 
=90
x , 
59
.
189 , 
270 or 
41
.
350
Solve 0
sin
2
sin =
− x
x
Solution
0
sin
2
sin =
− x
x
Example 32
Example 33

BARAKA LO1BANGUT1
0
sin
cos
sin
2 =
− x
x
x
( ) 0
1
cos
2
sin =
−
x
x
Case 1: 0
sin =
x
( )
0
sin 1
−
=
x

= 0
x , 
180 , 
360
Case 2: 0
1
cos
2 =
−
x
5
.
0
cos =
x
( )
5
.
0
cos 1
−
=
x

= 60
x or 
300
The solution set is 
= 0
x , 
60 , 
180 , 
300 and 
360
Solve 1
cot
2
cot
3 =
+ x
x from 



− 180
180 x
Solution
1
cot
2
cot
3 =
+ x
x
1
tan
1
2
tan
1
3 =
+






x
x
1
tan
1
tan
2
tan
3
3 2
=
+
−
x
x
x
x
x tan
2
2
tan
3
3 2
=
+
−
0
5
tan
2
tan
3 2
=
−
+ x
x
Example 34

BARAKA LO1BANGUT1
( )( ) 0
1
tan
5
tan
3 =
−
+ x
x
Case 1: 0
5
tan
3 =
+
x
3
5
tan −
=
x






−
= −
3
5
tan 1
x


−
= 96
.
120
,
04
.
59
x
Case 2: 1
tan =
x
( )
1
tan 1
−
=
x

= 45
x


−
= 96
.
120
,
45
,
04
.
59
x
EXERCISE 7
Solve the following from 


 360
0 
1. 3
sin
2
cos +
= 

2. ( )

+
= 10
sin
3
cos 

3. 1
sin
cos
2
sin 2
=
+ 


4. 1
2
cos
2
2
cos
3 2
=
+ 

5. 2
3
cot
3
cot
5 2
=
− 

6. 1
2
tan
tan
2 2
+
= 

7. 1
2
cos
2
sin +
= 

8. 
 cot
tan =
9. 4
2
sec4
=

10. 0
2
3
sin
5
3
sin
3 2
=
+
− 


BARAKA LO1BANGUT1
SOLUTION OF EQUATIONS OF THE FORM c
b
a =
+ θ
sin
θ
cos
(a) From the half angle formula, we derive the so called t formula,
( ) ( )
2
cos
2
sin
2
sin A
A
A =
( ) ( )
( ) ( )
2
sin
2
cos
2
cos
2
sin
2
sin 2
2 A
A
A
A
A
+
=
( )
( )
2
tan
1
2
tan
2
sin 2 A
A
A
+
=
Let ( )
2
tan A
t =
Therefore,
2
1
2
sin
t
t
A
+
=
(b) ( ) ( )
2
sin
2
cos
cos 2
2 A
A
A −
=
( ) ( )
( ) ( )
2
sin
2
cos
2
sin
2
cos
cos 2
2
2
2
A
A
A
A
A
+
−
=
( )
( )
2
tan
1
2
tan
1
cos 2
2
A
A
A
+
−
=
Let ( )
2
tan A
t =
Therefore, 2
2
1
1
cos
t
t
A
+
−
=

BARAKA LO1BANGUT1
(c)
( )
( )
2
tan
1
2
tan
2
tan 2 A
A
A
−
=
Let ( )
2
tan A
t =
Then,
2
1
2
tan
t
t
A
−
=
Solve 0
1
sin
cos
2 =
−
+ 
 from 


 360
0 
Solution
0
1
sin
cos
2 =
−
+ 

0
1
1
2
1
1
2 2
2
2
=
−






+
+








+
−
t
t
t
t
( ) 0
1
2
2
2 2
2
=
+
−
+
− t
t
t
0
1
2
3 2
=
−
− t
t
( )( ) 0
1
3
1 =
+
− t
t
1
=
t or
3
1
−
=
t
But ( )
2
tan 
=
t
Case 1: ( ) 1
2
tan =

( )
1
tan
2
1
−
=

 


= 225
,
45
2



= 450
,
90

Case 2: ( ) 3
1
2
tan −
=

( )
3
1
tan
2
1
−
= −

 


= 341.57
,
57
.
161
2

Example 35

BARAKA LO1BANGUT1


= 683.14
,
14
.
323

The solution is 

= 14
.
323
,
90

Note that, other solutions are out of range given.
Solve equation 2
sin
4
cos
3 =
+ 
 from 


 360
0 
Solution
2
sin
4
cos
3 =
+ 

2
1
2
4
1
1
3 2
2
2
=






+
+








+
−
t
t
t
t
2
2
2
2
8
3
3 t
t
t +
=
+
−
0
1
8
5 2
=
−
− t
t
7165
.
1
=
t or 1165
.
0
−
=
t
( ) 7165
.
1
2
tan =

( )
7165
.
1
tan
2
1
−
=

 


= 239.77
,
77
.
59
2



= 479.54
,
54
.
119

( ) 1165
.
0
2
tan −
=

( )
1165
.
0
tan
2
1
−
= −

Example 36

BARAKA LO1BANGUT1
 


= 36
.
353
,
36
.
173
2



= 706.72
,
72
.
346


= 72
.
346
,
54
.
119

Solve 1
sin
cos
3 =
+ 
 form 


 360
0 
Solution
1
sin
cos
3 =
+ 

1
1
2
1
1
3 2
2
2
=






+
+








+
−
t
t
t
t
( ) 2
2
1
2
1
3 t
t
t +
=
+
−
0
2
3
1
3 2
2
=
+
−
−
− t
t
t
( ) 0
1
3
2
1
3 2
=
−
+
+
+
− t
t
1
=
t or 2679
.
0
−
=
t
( ) 1
2
tan =

( )
1
tan
2
1
−
=

 


= 225
,
45
2



= 450
,
90

Example 37

BARAKA LO1BANGUT1
( ) 2679
.
0
2
tan −
=

( )
2679
.
0
tan
2
1
−
= −

 


= 345
,
165
2



= 690
,
330



=
 330
,
90

Solution of equations of the form c
b
a =
+ θ
sin
θ
cos :
(a) Expressing equation of the form c
b
a =
+ θ
sin
θ
cos into the form
( )
α
θ
cos
θ
sin
θ
cos −
=
+ R
b
a





 sin
sin
cos
cos
sin
cos R
R
b
a +
=
+


 cos
cos
cos R
a =

cos
R
a = (1)


 sin
sin
sin R
b =

sin
R
b = (2)

 2
2
2
2
2
2
sin
cos R
R
b
a +
=
+ (Square and add equation 1 and 2)
( )

 2
2
2
2
2
sin
cos +
=
+ R
b
a
The value of R is 2
2
b
a
R +
=


cos
sin
R
R
a
b
= (Divide equation 1 and 2)

BARAKA LO1BANGUT1
a
b
=

tan
Therefore, 





= −
a
b
1
tan

(b) EXPRESSING EQUATION OF THE FORM c
b
a =
+ θ
sin
θ
cos
INTO THE FORM ( )
α
θ
sin
θ
sin
θ
cos +
=
+ R
b
a
( )



 +
=
+ sin
sin
cos R
b
a





 sin
cos
cos
sin
sin
cos R
R
b
a +
=
+
Compare


 sin
cos
cos R
a =

sin
R
a =
(1)


 cos
sin
sin R
b =

cos
R
b =
(2)

 2
2
2
2
2
2
cos
sin R
R
b
a +
=
+ (Square and add equation 1 and 2)
( )

 2
2
2
2
2
cos
sin +
=
+ R
b
a
2
2
2
2
2
b
a
R
R
b
a +
=

=
+
b
a
R
R
=


cos
sin
(Divide equation 1 by 2)
b
a
=

tan

BARAKA LO1BANGUT1






= −
b
a
1
tan















+
+
=
+ −
b
a
b
a
b
a 1
2
2
tan
cos
sin
cos 


Express 
 sin
4
cos
3 + in the form ( )

 −
cos
R
Solution
Compare and expand ( )

 −
cos
R





 sin
sin
cos
cos
sin
4
cos
3 R
R +
=
+


 cos
cos
cos
3 R
=

cos
3 R
=
(1)


 sin
sin
sin
4 R
=

sin
4 R
=
(2)

 2
2
2
2
2
2
sin
cos
4
3 R
R +
=
+
1. Express c
b
a =
− 
 sin
cos in the form ( )

 −
sin
R
2. Express c
b
a =
− 
 sin
cos in the form ( )

 +
cos
R
Questions
Example 38

BARAKA LO1BANGUT1
( )

 2
2
2
sin
cos
16
9 +
=
+ R
2
25 R
= 5

=
 R


cos
sin
3
4
R
R
=
3
4
tan =
 

=






= −
1
.
53
3
4
tan 1

Therefore, ( )

−

=
+ 1
.
53
cos
5
sin
4
cos
3 


Express 39
cos
12
sin
5 =
− 
 in the form ( )

 −
sin
R
Solution
Compare and expand ( )



 −
=
− sin
cos
12
sin
5 R





 sin
cos
cos
sin
cos
12
sin
5 R
R −
=
−


 cos
sin
sin
5 R
=

cos
5 R
=
(1)


 sin
cos
cos
12 R
=

sin
12 R
=
(2)

 2
2
2
2
2
2
sin
cos
12
5 R
R +
=
+
( )

 2
2
2
sin
cos
144
25 +
=
+ R
2
169 R
= 13

=
 R
Example 40

BARAKA LO1BANGUT1
5
12
cos
sin
=


R
R
( )
4
.
2
tan
5
12
tan 1
−
=

=
 


= 4
.
67

( )

−

=
− 4
.
67
sin
13
cos
12
sin
5 

 but 39
cos
12
sin
5 =
− 

( ) 39
4
.
67
sin
13 =

−
 
( ) 3
4
.
67
sin 
=

−

Write 3
sin
cos
2 =
− 
 in the form ( )

 +
cos determine the
maximum and minimum value.
Solution
Compare and expand
( )



 +
=
− cos
sin
cos
2 R





 sin
sin
cos
cos
sin
cos
2 R
R −
=
−


 cos
cos
cos
2 R
=

cos
2 R
=


 sin
sin
sin R
=

sin
1 R
=
( ) 
 2
2
2
2
1
2
sin
cos
1
2 R
R +
=
+
( )

 2
2
2
sin
cos
1
2 +
=
+ R
Example 41

BARAKA LO1BANGUT1
3
2
=
R 3

=
 R
2
1
tan
cos
sin
=
= 


R
R

=






= −
3
.
35
2
1
tan 1

( ) 3
3
.
35
cos
3
sin
cos
2 =

+

=
− 


( ) 3
3
.
35
cos
3 =

+
 
Therefore, ( )
3
3
3
.
35
cos 
=

+
 , Maximum is
3
3
=
R and minimum
is
3
3
−
=
R
QUESTIONS
(a) Express each of the following in the form ( )

 
sin
R
1. 2
sin
3
cos =
+ 

2. 5
cos
4
sin
3 =
− 

3. 1
sin
5
cos
4 =
− 

4. 3
sin
9
cos
10 =
− 

(b) Express each of the following in the form ( )

 
cos
R
5. 
 sin
cos
2 −
6. 5
sin
10
cos
7 =
− 

7. 9
sin
cos
4 =
+ 

8. 4
cos
2
sin
3 =
+ 


BARAKA LO1BANGUT1
We now clearly know how to express the function c
b
a =
 
 sin
cos in
the form of ( )

 
cos
R or ( )

 
sin
R . This concept is useful in solving
trigonometric equation of the form c
b
a =
 
 sin
cos .
Solve 2
sin
4
cos
3 =
+ 
 for  from 


 360
0 
Solution
( )



 −
=
+ cos
sin
4
cos
3 R





 sin
sin
cos
cos
sin
4
cos
3 R
R −
=
+


 cos
cos
cos
3 R
=

cos
3 R
=


 sin
sin
sin
4 R
=

sin
4 R
=
3
4
tan
cos
sin
=
= 


R
R

=






= −
1
.
53
3
4
tan 1


 2
2
2
2
2
2
sin
cos
4
3 R
R +
=
+
5
25 2
=

= R
R
USING R-FORMULA TO SOLVE TRIGONOMETRIC
EQUATION OF THE FORM c
b
a =
+ θ
sin
θ
cos
Example 41

BARAKA LO1BANGUT1
( ) 2
1
.
53
cos
5
sin
4
cos
3 =

−
=
+ 


( )
5
2
1
.
53
cos =

−

( )
4
.
0
cos
1
.
53 1
−
=

−

 


=

− 6
.
293
,
4
.
66
1
.
53

Solution 

= 7
.
346
,
5
.
119

Find the maximum and minimum value of 
 cos
4
sin
2 − and the
corresponding values of  between 
0 to 
360 inclusive.
Solution
( )



 −
=
− sin
cos
4
sin
2 R





 sin
cos
cos
sin
cos
4
sin
2 R
R −
=
−


 cos
sin
sin
2 R
=

cos
2 R
=


 sin
cos
cos
4 R
=

sin
4 R
=
( )

 2
2
2
2
2
sin
cos
4
2 +
=
+ R
5
2
20 =
=
R
The minimum value is 5
2
− and the maximum value is 5
2
Example 42

BARAKA LO1BANGUT1



tan
cos
sin
2
4
=
=
R
R
( )
2
tan 1
−
=
 
=
 4
.
63

( )

− 4
.
63
sin
5
2 
At maximum, ( ) 5
2
4
.
63
sin
5
2 =

−

( )
1
sin
4
.
63 1
−
=

−
 
=

−
 90
4
.
63
 
=

+

=
 4
.
153
4
.
63
90

At minimum, ( ) 5
2
4
.
63
sin
5
2 −
=

−

( )
1
sin
4
.
63 1
−
=

− −
 
=


=

−
 4
.
333
270
4
.
63 

Therefore the solutions are ( )

4
.
153
,
5
2 and ( )

− 4
333
,
5
2 .
Solve 1
sin
5
cos =
+ 
 from 


 360
0 
Solution
( )



 −
=
+ cos
sin
5
cos R





 sin
sin
cos
cos
sin
5
cos R
R +
=
+


 cos
cos
cos R
= 1
cos =
 
R


 sin
sin
sin
5 R
= 
sin
5 R
=

( ) ( )

 2
2
2
2
2
sin
cos
5
1 +
=
+ R
6
6 2
=

= R
R
Example 43

BARAKA LO1BANGUT1
5
tan
5
cos
sin
=

= 


R
R
( ) 
=
= −
9
.
65
5
tan 1

( ) ( )

−
=
− 9
.
65
cos
6
cos 


R
( ) 1
9
.
65
cos
6 =

−







=

− −
6
1
cos
9
.
65 1


=

− 9
.
65
9
.
65

The solution, 

= 360
,
0

EXERCISE 8
(a) Solve each of the following using R-formula from 
0 to 
360 inclusive
1. 5
sin
2
cos
3 =
+ 

2. 3
cos
2
sin
2 =
+ 

3. 1
3
sin
3
cos =
+ 

4. 7
2
cos
12
2
sin
5 =
− 

5. ( ) ( ) 2
sin
3
cos 2
1
2
1 =
− x
x
6. 7
sin
3
cos
2 =
+ 

7. 1
sin
3
1
cos
4
1
=
+ 

8. 0
2
2
sin
3
2
cos =
+
+ 

9. ( ) 3
sin
3
cos
2 2
=
− 

10. ( ) 10
2
sin
5
2
cos
3 =
+ 

11. 4
cos
sin =


12. 0
3
cos
3
sin =



BARAKA LO1BANGUT1
13. 6
cos
12
sin
5 =
− 

14.
2
1
cos
sin =
+ 

15. 2
sin
7
cos =
− 

16. 4
cos
7
sin
2 =
+ 

17. 4
sec
2
tan
3 =
− 

18. 10
2
cos
15
cos
sin
4 =
+ 


19. 0
sin
2
cos =
+ 

20. 1
cos
sin
3 =
+ 

(b) Find the maximum and minimum value of the expressions below and find
the corresponding value of 
21. 
 sin
cos +
22. 
 sin
3
cos
2 +
23. 
 sin
cos
2 −
24. 
 sin
5
cos −
25. 
 sin
3
cos
2 −
26. 
 sin
2
cos
3 −

BARAKA LO1BANGUT1
The solution of trigonometric equations are divided into two major parts
1. Principal solution, are the solution which take an angle form 
0 to 
360
inclusively.
2. General solution, these are all possible solution of the trigonometric
equations
(a) If α
θ sin
sin = , then ( ) α
1
πn n
−
+
=
θ
(b) If α
θ cos
cos = , then α
πn
θ 
= 2
(c) If α
θ tan
tan = , then α
πn+
=
θ
For all three cases above, 

n (n is any integer)
Find the general solution of 1
sin −
=

Solution
Given 1
sin −
=

( )
1
sin
1
sin 1
−
=

−
= −


2
π
90 −
=

−
=

( ) α
1
-
πn
θ n
+
=
The general solution is ( ) 





−
=
2
1

 n
-
πn
tan =
x
Solution
GENERAL SOLUTION TO THE TRIGONOMETRIC
EQUATIONS
Example 44
Example 45

BARAKA LO1BANGUT1
3
tan =
x
( )
3
tan 1
−
=
x
3

=
x
The general solution is
3
2


 +
= n , for Z
n
Find the general solution
2
1
4
sin =






+

x
Solution
2
1
4
sin =






+

x






=
+ −
2
1
sin
4
1

x
4
4


=
+
x
0
=
x
The general solution is n

 =
sin
3
cos =
+ x
x
Solution
Example 46
Example 47

BARAKA LO1BANGUT1
2
sin
3
cos =
+ x
x
( )

−
=
+ x
R
x
x cos
sin
3
cos

 sin
sin
cos
cos
sin
3
cos x
R
x
R
x
x +
=
+

cos
cos
cos x
R
x =
1
cos =

R

sin
sin
sin
3 x
R
x =
3
sin =

R
2
2
3
1 +
=
R  10
=
R
( )
3
tan
3
tan 1
−
=

= x


= 56
.
71

( )

−
=
+ 56
.
71
cos
10
sin
3
cos x
x
x
( ) 2
56
.
71
cos
10 =

−
x
( )
10
2
56
.
71
cos =

−
x






=

− −
10
2
cos
56
.
71 1
x
The general solution is 

=

− 77
.
50
2
56
.
71 n
x 

+
= 33
.
122
2 n
x  or 
−
= 79
.
20
2 n
x 

BARAKA LO1BANGUT1
EXERCISE 9
Find the general solution of the following trigonometric equations
1.
2
1
sin −
=
x
2.
2
1
2
cos =
x
3. 1
tan −
=
x
4.
2
1
3
sin −
=






−

x
5. 





=






+
3
tan
3
tan


x
6. 3
tan =
x
7.
2
1
sin2
=
x
8.
2
3
2
cos =






−

x
9.
2
3
4
cos −
=






+

x
10. 2
sin
2
cos =
+ x
x
Consider the sine compound angles
( )
( )



−
=
−
+
=
+
+
B
A
B
A
B
A
B
A
B
A
B
A
sin
cos
cos
sin
sin
sin
cos
cos
sin
sin
( ) ( ) B
A
B
A
B
A cos
sin
2
sin
sin =
−
+
+
FACTOR FORMULA

BARAKA LO1BANGUT1
Let C
B
A =
+ and D
B
A =
−



=
−
=
+

D
B
A
C
B
A
Then
2
D
C
A
+
= and
2
D
C
B
−
=
The equation become, 




 −





 +
=
+
2
cos
2
sin
2
sin
sin
D
C
D
C
D
C
Also, from
( )
( )



−
=
−
+
=
+
−
B
A
B
A
B
A
B
A
B
A
B
A
sin
cos
cos
sin
sin
sin
cos
cos
sin
sin
( ) ( ) B
A
B
A
B
A sin
cos
2
sin
sin =
−
−
+
Substituting the values of B
A+ and B
A− ,





 −





 +
=
−
2
sin
2
cos
2
sin
sin
D
C
D
C
D
C
Consider the cosine compound angles
( )
( )



+
=
−
−
=
+
+
B
A
B
A
B
A
B
A
B
A
B
A
sin
sin
cos
cos
cos
sin
sin
cos
cos
cos
( ) ( ) B
A
B
A
B
A cos
cos
2
cos
cos =
−
+
+





 −





 +
=
+
2
cos
2
cos
2
cos
cos
D
C
D
C
D
C
Also, from
( )
( )



+
=
−
−
=
+
−
B
A
B
A
B
A
B
A
B
A
B
A
sin
sin
cos
cos
cos
sin
sin
cos
cos
cos
( ) ( ) B
A
B
A
B
A sin
sin
2
cos
cos −
=
−
−
+

BARAKA LO1BANGUT1





 −





 +
−
=
−
2
sin
2
sin
2
cos
cos
D
C
D
C
D
C
PROVING IDENTITIES INVOLVING FACTOR FORMULA
Prove that 




 −





 +
=
+
−
2
tan
2
cot
sin
sin
sin
sin C
B
C
B
C
B
C
B
Solution





 −





 +





 −





 +
=
+
−
2
cos
2
sin
2
2
sin
2
cos
2
sin
sin
sin
sin
C
B
C
B
C
B
C
B
C
B
C
B
FACTOR FORMULA SUMMARY
Sine





 −





 +
=
+
2
cos
2
sin
2
sin
sin
D
C
D
C
D
C and





 −





 +
=
−
2
sin
2
cos
2
sin
sin
D
C
D
C
D
C
Cosine





 −





 +
=
+
2
cos
2
cos
2
cos
cos
D
C
D
C
D
C and





 −





 +
−
=
−
2
sin
2
sin
2
cos
cos
D
C
D
C
D
C
Example 48

BARAKA LO1BANGUT1

















 −





 −

















 +





 +
=
2
cos
2
sin
2
sin
2
cos
C
B
C
B
C
B
C
B





 −





 +
=
2
tan
2
cot
C
B
C
B
hence shown
Prove that ( )
1
cos
2
2
sin
3
sin
2
sin
sin +
=
+
+ A
A
A
A
A
Solution
( )
A
A
A 3
sin
sin
2
sin +
+





 −





 +
+
2
3
cos
2
3
sin
2
2
sin
A
A
A
A
A
A
A
A cos
2
sin
2
2
sin +
( )
A
A cos
2
1
2
sin +
Prove that A
A
A
A sin
3
cos
2
4
sin
2
sin −
=
−
Solution
A
A
A
A sin
3
cos
2
4
sin
2
sin −
=
−





 −





 +
=
−
2
4
2
sin
2
4
2
cos
2
4
sin
2
sin
A
A
A
A
A
A
( ) ( )
A
A −
= sin
3
cos
2
A
Asin
3
cos
2
−
=
Remember that
( ) ( )
A
A cos
cos =
−
Example 49
Example 50

BARAKA LO1BANGUT1
Show that 





=
+
+
−
+
2
tan
cos
sin
1
cos
sin
1 A
A
A
A
A
(not concerned with factor
formula)
Solution
( )
( ) A
A
A
A
A
A
A
A
sin
cos
1
sin
cos
1
cos
sin
1
cos
sin
1
+
+
+
−
=
+
+
−
+


















+












−






+


















+












+






−
=
2
cos
2
sin
2
2
sin
2
cos
1
2
cos
2
sin
2
2
sin
2
cos
1
2
2
2
2
A
A
A
A
A
A
A
A


















+






























+












=
2
cos
2
sin
2
2
cos
2
2
cos
2
sin
2
2
sin
2
2
2
A
A
A
A
A
A
















































=
2
cos
2
sin
2
cos
2
2
cos
2
sin
2
sin
2
A
A
A
A
A
A






=












=
2
tan
2
cos
2
sin
A
A
A
Hence shown
Example 51

BARAKA LO1BANGUT1
Prove that ( ) ( ) A
A
A cos
150
sin
150
sin =
−

+
+

Solution
Left hand side
( ) ( ) A
A
A cos
150
sin
150
sin =
−

+
+

( ) ( ) ( ) ( )





 −

−
+






 −

+
+

2
150
150
cos
2
150
150
sin
2
A
A
A
A











 
=
2
2
cos
2
300
sin
2
A
( ) A
cos
150
sin
2 
=
A
A cos
cos
2
1
2 =






= Hence shown.
Prove that A
A
A
A
A
tan
5
cos
3
cos
3
sin
5
sin
=
+
−
Solution





 −





 +





 −





 +
=
+
−
2
5
3
cos
2
5
3
cos
2
2
3
5
sin
2
3
5
cos
2
5
cos
3
cos
3
sin
5
sin
A
A
A
A
A
A
A
A
A
A
A
A
( ) ( )
( ) ( )
A
A
A
A
A
A
A
tan
cos
sin
cos
4
cos
sin
4
cos
=
=
=
Example 52
Example 53

BARAKA LO1BANGUT1
If π
C
B
A =
+
+ , show that


















+
=
+
+
2
sin
2
sin
2
sin
4
1
cos
cos
cos
C
B
A
C
B
A
Solution














−






+





 −





 +
=
+
+
2
sin
2
cos
2
cos
2
cos
2
cos
cos
cos 2
2 C
C
B
A
B
A
C
B
A














−
+





 −





 +

2
sin
2
1
2
cos
2
cos
2 2 C
B
A
B
A
But
2
π
2
C
B
A −
=
+






−
+





 −





 −

2
sin
2
1
2
cos
2
π
cos
2 2 C
B
A
C






−





 −






−
+

2
sin
2
2
cos
2
2
π
cos
2
1 2 C
B
A
C






−





 −






+

2
sin
2
2
cos
2
sin
2
1 2 C
B
A
C












−





 −






+

2
sin
2
cos
2
sin
2
1
C
B
A
C
But 




 +
−
=

=
+
+
2
2
π
2
2
π
2
2
B
A
C
C
B
A
Example 54

BARAKA LO1BANGUT1



















 +
−
−





 −






+

2
2
π
sin
2
cos
2
sin
2
1
B
A
B
A
C











 +
−





 −






+

2
cos
2
cos
2
sin
2
1
B
A
B
A
C












−






−






+

2
sin
2
sin
2
2
sin
2
1
B
A
C
























+

2
sin
2
sin
2
2
sin
2
1
B
A
C


















+

2
sin
2
sin
2
sin
4
1
C
B
A
SIMPLIFYING TRIGONOMETRIC IDENTITIES INVOLVING
FACTOR FORMULA
Simplify
A
A
A
A
A
A
A
A
4
sin
3
sin
cos
2
cos
3
cos
6
sin
cos
8
sin
−
−
Solution
( ) ( )
  ( ) ( )
 
( ) ( )
  ( ) ( )
 
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
A
4
3
cos
4
3
cos
2
1
2
cos
2
cos
2
1
3
6
sin
3
6
sin
2
1
8
sin
8
sin
2
1
−
−
+
+
−
+
+
−
+
+
−
−
+
+
=
A
A
A
A
A
A
A
A
cos
7
cos
cos
3
cos
3
sin
9
sin
7
sin
9
sin
−
+
+
−
−
+
=
A
A
A
A
3
cos
7
cos
3
sin
7
sin
+
−
=
Example 55

BARAKA LO1BANGUT1





 −





 +





 −





 +
=
2
3
7
cos
2
3
7
cos
2
2
3
7
sin
2
3
7
cos
2
A
A
A
A
A
A
A
A
A
A
A
2
tan
2
cos
2
sin
=
=
Simplify ( ) ( )2
2
sin
sin
cos
cos B
A
B
A +
+
+
Solution
( ) ( )2
2
sin
sin
cos
cos B
A
B
A +
+
+
( ) ( )2
2
sin
sin
cos
cos B
A
B
A +
+
+
2
2
2
cos
2
sin
2
2
cos
2
cos
2 












 −





 +
+













 −





 +
=
B
A
B
A
B
A
B
A





 −





 +
+





 −





 +
=
2
cos
2
sin
4
2
cos
2
cos
4 2
2
2
2 B
A
B
A
B
A
B
A











 +
+





 +





 −
=
2
sin
2
cos
2
cos
4 2
2
2 B
A
B
A
B
A





 −
=
2
cos
4 2 B
A
Example 56

BARAKA LO1BANGUT1
EXERCISE 10
1. Prove that
( ) ( )
( ) ( )
A
A
A
A
A
tan
2
sin
4
sin
2
cos
4
cos
−
=
+
−
2. Show that A
A
A
A
A
3
tan
2
cos
4
cos
2
sin
4
sin
=
+
+
3. Show that
( )
( )
B
A
B
A
B
A
B
A
+
−
=
+
−
2
1
2
1
tan
tan
sin
sin
sin
sin
4. Show that 
=

+
 5
cos
2
40
sin
50
sin
5. Prove that ( )
B
A
B
A
B
A
+
=
+
+
2
1
tan
cos
cos
sin
sin
6. Show that ( )
A
A
A
A
A 5
cos
3
cos
cos
2
16
1
sin
cos 2
3
−
−
=
7. Prove that ( ) ( ) ( ) ( ) ( ) ( )
A
A
A
A
A
A 3
cos
2
cos
cos
4
6
cos
4
cos
2
cos
1 =
+
+
+
8. Show that ( )




 cos
2
1
2
sin
3
sin
2
sin
sin +
=
+
+
9. Simplify








9
cos
7
cos
5
cos
3
cos
9
sin
7
sin
5
sin
3
sin
+
+
+
+
+
+
10. Simplify (a)
A
A
A
cos
2
sin
4
sin +
(b)
A
A
A
3
sin
2
2
9
cos
2
3
cos 





−






11. Simplify
A
A
A
A
2
sin
5
cos
3
sin
7
sin −
12. Prove that 3
50
sin
70
sin
70
cos
50
cos
=

−


−

13. Simplify

+


+

50
sin
70
sin
70
cos
50
cos

BARAKA LO1BANGUT1
5
sin
3
sin
sin =
+
+ 


Solution
0
5
sin
3
sin
sin =
+
+ 


( ) 0
3
sin
5
sin
sin =
+
+ 


0
3
sin
2
5
cos
2
5
sin
2 =
+





 −





 +





0
3
sin
2
cos
3
sin
2 =
+ 


( ) 0
1
2
cos
2
3
sin =
+


0
3
sin =

( )
0
sin
3 1
−
=

( ) 
 n
1
πn
3 −
+
=
πn
3 =

3
πn
=
 
2
1
2
cos −
=

( )
5
.
0
cos
2 1
−
= −

3
2
120
2

 =

=


 
= n
2
2
Example 57
SOLVING TRIGONOMETRIC INVOLVING FACTOR FORMULA

BARAKA LO1BANGUT1
3


 
= n
The general solution set is
3


 
= n or
3
πn
=

5
sin
sin =
+ 

Solution
Given 0
5
sin
sin =
+ 

0
2
5
cos
2
5
sin
2
5
sin
sin =





 −





 +
=
+






0
2
cos
3
sin
2 =
= 

Case 1: 0
3
sin =

( )
0
sin
3 1
−
=
= 

 n
=
3
3


n
=
Case 2: 0
2
cos =

( )
0
cos
2 1
−
=

2
2

 =
2
2
2


 
= n
Example 58

BARAKA LO1BANGUT1
The general solution set is
4


 
= n
Find the general solution of ( ) ( ) 0
40
cos
20
cos =

−
−

+ x
x
Solution
( ) ( ) 0
40
cos
20
cos =

−
−

+ x
x
( ) ( ) 0
2
60
sin
2
20
2
sin
2
40
cos
20
cos =





 





 
−
−
=

−
−

+
x
x
x
( ) 0
10
sin
2
1
2 =

−






−
= x
( ) 0
10
sin =

−
x
( )
0
sin
10 1
−
=

−
x

n
x =

−10
18

 +
= n
x
Find the general solution of ( ) ( ) 0
20
2
cos
70
2
cos =

−
+

+ 

Solution
( ) ( ) 0
20
2
cos
70
2
cos =

−
+

+ 

Example 59
Example 60

BARAKA LO1BANGUT1
( ) ( ) 




 





 
−
−
=

−
+

+
2
90
sin
2
50
4
sin
2
40
2
cos
20
2
cos



( ) 0
25
2
sin
45
sin
2 =

−

− 
( ) 0
25
2
sin =

−

( )
0
sin
25
2 1
−
=

−


 n
=

−25
2

+
= 25
2 
 n
72
5
2


 +
=
n
EXERCISE 11
1. Given that x
B
A =
+ 2
sin
2
sin and y
B
A =
+ 2
cos
2
cos , show that
( )
B
A
y
x +
= tan . Prove that
( )
B
A
B
A
y
y
x
x
cos
cos
sin
2
4
2
2
+
=
+
+
and deduce an
expression for B
Atan
tan in terms of x and y
2. Show that if
y
y
+
−
=
1
1
sin then y
=






−
2
4
tan


3. If 2
3
tan =
A simplify
A
A
A
A
A
A
5
cos
3
cos
cos
5
sin
3
sin
sin
+
+
+
+
4. If p
B
A =
+sin
sin and q
B
A =
+ cos
cos , show that
(a) ( ) 2
2
2
sin
q
p
pq
B
A
+
=
+

BARAKA LO1BANGUT1
(b) ( ) 2
2
2
2
cos
p
q
p
q
B
A
+
−
=
+
(c) 2
2
3
8
tan
tan
p
q
pq
B
A
−
=
+
(d)
( )( )
2
2
2
2
2
2
2
2
cos
2
cos
q
p
q
p
p
q
B
A
+
−
+
−
=
+
5. Find the general solution of each of the following
(a) 0
cos
3
cos =
+ 

(b) 

 6
cos
4
cos
2
cos =
+
(c) ( ) ( ) 0
10
3
sin
10
3
sin =

−
+

+ 

(d) 0
5
cos
cos =
− 

(e) 0
5
sin
3
sin
sin =
+
+ 


(f) 0
cos
cos
2
cos 2
5
2
3
2
1 =
+
+ 


An angle can be measured in degree or radians.
Consider the diagrams below
RADIANS AND SMALL ANGLES

r
r
L
O
Using ration theorem

=
360

c
l
But r
c 
2
=

=
360
2

r
l

r
l 2
360 =


=
180

r
l

=
180
rad
r


BARAKA LO1BANGUT1

=
=
180
rad
r
r
l 
The radian formula
Express the following in radians
(a) 
45 (b) 
60 (c) 
70
Solution
(a)

=
180
rad
r

( )
4
180
45
45


=


=

(b)
( )
3
180
60
60


=


=

(c)
( )
18
7
180
70
70


=


=

Express the following in degree
(a)
3
2
(b)
8

(c)
7
5
Solution

rad
180
deg

=
(a) 
=











 
= 120
3
2
180
deg


Example 61
Example 62

BARAKA LO1BANGUT1
(b) 
=











 
= 5
.
22
8
180
deg


(c) '
34
128
7
5
180
deg 
=











 
=


SMALL ANGLES
Let refer the Maclaurin’s series for sine, cosine and tangent, through these
series we deduce the approximations for small angles as 0
→
 .
Sine
...
!
7
!
5
!
3
sin
7
5
3
+
−
+
−
=





( ) 







+
−
+
−
=
→
→
...
!
7
!
5
!
3
lim
sin
lim
7
5
3
0
0







Ignoring third and higher derivatives, we get
( ) 



→
sin
lim
0
Cosine
...
!
6
!
4
!
2
1
cos
6
4
2
+
−
+
−
=




( ) 







+
−
+
−
=
→
→
...
!
6
!
4
!
2
1
lim
cos
lim
6
4
2
0
0






Ignoring the higher powers
( ) 



→
sin
lim
0

BARAKA LO1BANGUT1
( )
!
2
1
cos
lim
2
0



−

→
Tangent
...
315
17
15
2
3
1
tan 7
5
3
+
+
+
+
= 




( ) 





+
+
+
+
=
→
→
...
315
17
15
2
3
1
lim
tan
lim 7
5
3
0
0







Ignoring the higher power
( ) 



→
tan
lim
0
Simplify



tan
sin
4
cos
1−
if  is small
Solution
( )
8
2
16
2
4
1
1
tan
sin
4
cos
1
2
2
2
2
=
=








−
−
=
−







( )
2
1
cos
lim
2
0



−

→
( ) 



→
tan
lim
0
Example 63

BARAKA LO1BANGUT1
Evaluate the following as 0
→

(a)



2
cos
1
tan
−
(b)
( )




sin
sin
sin +
+
Solution
(a)
( ) 2
1
2
2
2
1
1
2
cos
1
tan
lim 2
2
2
2
0
=
=








−
−
=






−
→ 







2
1
2
cos
1
tan
lim
0







−
→ 



(b)
( ) 
















 +
=





 +
+
→ sin
sin
cos
cos
sin
sin
sin
sin
lim
0




























−
+
=
sin
2
1
cos
2
( ) 






cos
sin
sin
sin
lim
0






 +
+
→
Example 64

BARAKA LO1BANGUT1

BARAKA LO1BANGUT1
As other functions, in trigonometric functions we can also determine domain
and range of a function. Were x being values which gives out an output of
the function.
Consider the function ( ) x
x
f cos
=
Domain of ( ) x
x
f cos
= is the set of all real numbers (x is in radian) and
range is any number in the intervals, 1
1 

− y thus 1

y . The domain
and range of cosine are the same for sine.
For tangent, domain are all values except ( )
2
1
2

+
n , where n is any integer,
range is a set of all real numbers.
Graph of sine and cosine
TRIGONOMETRIC FUNCTION
x
y sin
=
x
y cos
=

BARAKA LO1BANGUT1
Graph of tangent
Asymptote
Asymptotes, ( ) 2
π
1
2 +
= n
x

BARAKA LO1BANGUT1
Graph of secant
Asymptotes, ( ) 2
π
1
2 +
= n
x
Asymptote

BARAKA LO1BANGUT1
Graph of cosecant
Asymptotes, π
=
x Asymptote

BARAKA LO1BANGUT1
Graph of cotangent
Determine the domain and range of x
y cos
2
3+
= hence sketch the graph.
Solution
Using the standard domain and range
Range: 1
1 

− y this is the standard range
x
y cos
2
3+
=
1
cos
1 

− x
Multiply by 2 through out
2
cos
2
2 

− x
Asymptote
Asymptotes, π
=
x
Example 65

BARAKA LO1BANGUT1
Add 3 both sides
3
2
cos
2
3
3
2 +

+

+
−
The range becomes
5
1
5
cos
2
3
1 



+
 y
x
The range set is 5
1 
 y
Domain  


= x
x:
Determine the domain and range of the function ( ) x
x
f 2
sin
4+
−
=
Solution
Given, ( ) x
x
f 2
sin
4+
−
=
Domain  


= x
x:
Graph of x
y cos
2
3+
=
Example 66

BARAKA LO1BANGUT1
Range
Standard range, 1
1 

− y
( ) x
x
f
y 2
sin
4+
−
=
=
1
2
sin
1 

− x
Subtract 4 both sides
4
1
2
sin
4
4
4 −

+
−

−
− x
3
sin
4
5 −

+
−

− x
3
5 −


− y
Range  
3
5
: −


−
= y
y
( ) x
x
f 2
sin
4+
−
=

BARAKA LO1BANGUT1
Sketch the graph of ( ) x
x
x
f cos
sin +
=
Solution
Table of values
x -4 -3 -2 -1 0 1 2 3 4
y 0.10 -1.13 -1.33 -0.30 1.00 1.38 0.49 -0.85 -1.41
( ) x
x
f sin
=
Example 67

BARAKA LO1BANGUT1
PROVING EQUATION INVOLVING INVERSE OF
TRIGONOMETRIC FUNCTIONS
Prove that ( )
3
tan
5
3
tan
2
3
tan 1
1
1 −
−
−
=








+








Solution
Given ( )
3
tan
5
3
tan
2
3
tan 1
1
1 −
−
−
=








+








Let
2
3
tan
2
3
tan 1
=









= −
A
A and
5
3
tan
5
3
tan 1
=









= −
B
B
Therefore, ( )
3
tan 1
−
=
+ B
A
Introducing tan
( )
B
A
B
A
B
A
tan
tan
1
tan
tan
tan
−
+
=
+
( )
















−








+








=
+
5
3
2
3
1
5
3
2
3
tan B
A
( ) 3
7
10
10
3
7
10
7
10
3
7
10
3
1
10
3
2
3
5
tan =

=
=
−
+
=
+ B
A
( ) 3
tan =
+ B
A
Example 68

BARAKA LO1BANGUT1
Then ( )
3
tan 1
−
=
+ B
A
But A and B are








= −
2
3
tan 1
A and








= −
5
3
tan 1
B
Hence, ( )
3
tan
5
3
tan
2
3
tan 1
1
1 −
−
−
=








+








Prove that 





=






+





 −
−
−
65
63
sin
5
4
sin
13
5
sin 1
1
1
Solution
Let
13
5
sin
13
5
sin 1
=

=






−
A
A and
5
4
sin
5
4
sin 1
=

=






−
B
B






=
+ −
65
63
sin 1
B
A
( ) B
A
B
A
B
A sin
cos
cos
sin
sin +
=
+
( ) B
A
B
A
B
A sin
sin
1
sin
1
sin
sin 2
2
−
+
−
=
+
( ) 





















−
+
















−






=
+
5
4
13
5
1
5
4
1
13
5
sin
2
2
B
A
( ) 











+












=
+
13
12
5
4
5
3
13
5
sin B
A
( )
65
48
65
15
sin +
=
+ B
A
Example 69

BARAKA LO1BANGUT1
( )
65
63
sin =
+ B
A






=
+ −
65
63
sin 1
B
A
But A
=






−
13
5
sin 1
and B
=






−
5
4
sin 1






=






+





 −
−
−
65
63
sin
5
4
sin
13
5
sin 1
1
1
Hence shown.
SOLVING EQUATIONS INVOLVING INVERSES OF
TRIGONOMETRIC FUNCTIONS
Solve for x from ( ) 2
2
cos
cos 1
1 
=
+ −
−
x
x
Solution
( ) 2
2
cos
cos 1
1 
=
+ −
−
x
x
Let x
A
x
A =

= −
cos
cos 1
and ( ) 2
cos
2
cos 1
x
B
x
B =

= −
Then
2

=
+ B
A
( ) 





=
+
2
cos
cos

B
A
( ) 0
sin
sin
cos
cos
cos =
−
=
+ B
A
B
A
B
A
Example 70

BARAKA LO1BANGUT1
0
cos
1
cos
1
cos
cos 2
2
=





 −





 −
− B
A
B
A
( )( ) 0
2
1
1
2 2
2
=





 −





 −
− x
x
x
x
( )( ) 0
2
1
1
2 2
2
2
=
−
−
− x
x
x
( )( )
2
2
2
2
1
1
2 x
x
x −
−
=
( )( )
2
2
4
2
1
1
2 x
x
x −
−
=
4
2
2
4
2
2
1
2 x
x
x
x +
−
−
=
0
1
3 2
=
−
x
3
3
3
1

=

=
x
Solve for x from
2
4
tan
2
tan 1
1 
=
+ −
−
x
x
Solution
2
4
tan
2
tan 1
1 
=
+ −
−
x
x
Let x
A
x
A 2
tan
2
tan 1
=

= −
and x
B
x
B 4
tan
4
tan 1
=

= −
2

=
+ B
A
Example 71

BARAKA LO1BANGUT1
( ) 





=
+
2
tan
tan

B
A
3
tan
tan
1
tan
tan
=
−
+
B
A
B
A
( )( )
3
4
2
1
4
2
=
−
+
x
x
x
x
3
8
1
6
2
=
− x
x
2
3
8
3
6 x
x −
=
0
3
6
3
8 2
=
−
+ x
x
The solution is 198
.
0
=
x or 631
.
0
−
=
x

BARAKA LO1BANGUT1
1. Draw the graph of x
y cos
4
2+
= and state domain and range of y.
2. Evaluate the following
(a) 




 −
→ 


 5
sin
1
2
cos
lim
0
(b) 




 +
→ 


 2
5
tan
3
sin
lim
0
(c) 





+
−
+
→ 


 2
sin
1
cos
20
tan
7
21
lim
0
(d) 





→ 

 sin
4
lim
0
(e) 





→ 

 sin
2
tan
lim
0
(f) 





−
→ 

 2
cos
1
sin
lim
0
(g) 




 +
→ 


 2
cos
tan
3
sin
lim
0
(h) ( )

+
→
45
sin
lim
0


(i) ( )

+
→
30
cos
lim
0


(j) 





−
+
+
→ 


 cos
3
tan
3
5
sin
1
lim
0
3. Prove the following
(a) 





=






−





 −
−
−
425
297
cos
25
7
cos
17
15
cos 1
1
1
(b)







 −
=






−





 −
−
−
6
6
2
1
sin
3
1
cos
2
1
sin 1
1
1
MISCELLANEOUS EXERCISE

BARAKA LO1BANGUT1
(c) 





=






+





 −
−
−
36
77
tan
8
15
cot
4
3
tan 1
1
1
(d) 





=






−






− −
−
−
16
63
tan
13
5
cos
5
3
sin 1
1
1
(e) 





−
=






+





 −
−
−
204
253
tan
13
12
cos
5
4
sin
2 1
1
1
(f) 





=






+





 −
−
−
2
1
12
tan
5
7
tan
5
3
tan 1
1
1
(g) 





=






−





 −
−
−
325
323
sin
13
12
cos
5
4
sin
2 1
1
1
4. Solve the following
(a) 
=






+





 −
−
45
3
1
tan
2
1
tan 1
1
(b)
2
5
3
tan
5
4
sin 1
1 
=






+





 −
−
(c)
4
7
1
tan
3
4
tan 1
1 
=






−





 −
−
(d)
4
7
1
tan
3
1
tan
2 1
1 
=






+





 −
−
(e)
3
4
1
tan
13
12
cos 1
1 
=






+





 −
−
5. Find the general solution of the following
(a) 0
1
sin
3
2
cos =
+
− x
x
(b) 2
cos
4
sin
3 =
+ x
x
(c)
2
1
cos
sin =
x
x
(d) ( ) 1
cos
sin 2
1 =
+ x
x
(e) ( )( ) 0
1
sin
2
1
tan =
+
− x
x
6. Prove the following

BARAKA LO1BANGUT1
(a) x
x cos
2
8
cos
2
2
2
2 =
+
+
+
(b) 





= −
−
x
x
1
cos
sec 1
1
(c) 





= −
−
x
x
1
sin
csc 1
1
(d) 





= −
−
x
x
1
tan
cot 1
1
(e) x
x
x 2
2
2
sin
2
1
cot
1
1
cot
−
=
+
−
(f) x
x
x
x
x
x
sec
csc
cos
sin
sin
cos 2
2
−
=
−
(g)
A
A
A
A
cot
1
csc
1
csc
cot −
=
−
7. If π
=
+
+ C
B
A show that
C
B
A
C
B
A cos
cos
cos
2
1
cos
cos
cos 2
2
2
−
=
+
+
8. Simplify
(a)


−

+

30
tan
15
tan
1
30
tan
15
tan
(b) 

+

 40
cos
50
sin
50
cos
40
sin
9. Solve for x: ( ) ( )

+
=

+ 30
cos
30
sin x
x from 
0 to 
360 inclusive
10. Prove that
(a)
x
x
x
x
x
2
cos
cos
3
1
2
sin
sin
3
tan
+
+
+
=
(b) A
A
A
A
A
sin
cos
sin
cos
2
cos
+
=
−
11. Solve for  between 



− 180
180  from 1
6
tan
3 =

12. Draw the graph of
(a) x
y sin
= from 


 180
0 x
(b) x
y tan
= from
2
2




− x

BARAKA LO1BANGUT1
13. Without using the table find the value of
(a) 
18
sin ,
(b) 
18
cos
(c) 
18
tan
Hint: let 
=18
x , 
= 36
2x then 
= 90
5x ]
14. Prove the identity
(a) ( )





 4
sin
2
sin
cos
2
5
sin
3
sin
2
sin +
=
+
+
(b) Given that 1
sec
tan
3 =
− 
 find the possible value of 
 tan
sec
3 +
15. If 
=
+ 45
2 y
x show that








−
+
−
−
= −
x
x
x
x
y 2
2
1
tan
tan
2
1
tan
tan
2
1
tan
16. Evaluate 













→ x
x
x
1
sin
lim
0
17. Simplify
(a) ( )
x
x 1
1
cos
sin
sin −
−
+
(b)
















−





 −
−
2
3
tan
2
1
cos
tan 1
1
18. Use t substitution, to solve 0
2
cos
4
sin
3 =
−
+ x
x for x between 0 and
2π inclusive.
19. Prove that for any triangle
(a) ( ) A
a
c
b
C
B 2
1
2
1 cos
sin
−
=
−
(b) ( ) A
C
B
a
c
b 2
1
2
1 csc
cos −
=
+ , where a, b and c are sides of the
triangle.
20. Prove that
(a)
A
A
A
A
A
A
A
A
csc
sec
cot
tan
cot
tan
csc
sec
+
+
=
−
−
(b)
( )
B
B
A
A
A
A
A
2
sin
cos
2
cos
sin
sin
cos +
=
−
21. Solve the trigonometric equation 0
1
tan
sec2
=
−
+ 
 for 
 2
0 

22. Factorize 


 7
cos
5
cos
3
cos
cos +
−
−

BARAKA LO1BANGUT1
23. Prove that





cos
1
cos
2
sin
2
cos
3
cos
2
2
+
−
=
+
−
24. Show that A
A
A
A
A
A
A
A
A
5
tan
2
cos
sin
6
cos
3
sin
2
sin
sin
6
sin
3
sin
=
+
+
25. Show that






2
2
2
2
sin
sin
1
cos
cos
1
sin
2
sin
+
−
=
+
+
26. (a) Prove that
cos2θ
1
sin2θ
θ
tan
+
= and hence obtain 
15
tan and 
2
1
67
tan
as surds, in their simplest forms.
(b) Prove that A
A
A sin
1
cos
sin 2
1
2
1 −

=
− determining for what range
of angles A between 0 and 4π the positive sign is to be taken.
27. A line AP is drawn through the vertex A of a triangle ABC, so that P and C
are on opposite sides of AB, and angle θ
=
BAP . By projecting the sides
of the triangle ABC on AP, prove that
( ) ( )
θ
cos
θ
cos
θ
cos +
+
−
= A
b
B
a
c . Deduce from this result the two
formulae, A
b
B
a
c cos
cos +
= and A
b
B
a sin
sin = . Further, by
considering the special case B
A=
=
 , deduce that
1
cos
2
2
cos 2
−
= A
A A.E.B
28. (a) Find the solution between 
0 and 
360 inclusive of the equations
i) 
= 171
tan
2
3
cot
x
ii) 3
sin
2
cos
3
3
sin =
+
+ x
x
x
(b) Find the solution in radians between 0 and 2π of the equation
x
x sin
3
cos =
29. From a point A of a straight level road which runs due north, a peak bears
due west. A man walks along the road, starting from A and going at a
steady speed. After 10 min. the elevation of the peak is  ; after a further
20 min. it is  . If the elevation of the peak from A is  , prove that


 2
2
2
cot
cot
9
cot
8 −
= . If the man’s speed is 4m per hour and  ,

BARAKA LO1BANGUT1
 are '
18
17 and '
24
10 respectively, calculate the height of the peak
above the level of the road, to the nearest 50 m.
30. (a) Show that 3
tan =
A if ( )
A
A
A
A cos
sin
2
cos
sin −
=
+ , hence solve
( ) 0
cos
sin
2
cos
sin =
−
−
+ A
A
A
A for 


 360
0 A
(b) A function f is defined by x
x
f sin
2
3
)
( −
= for 


 360
0 x find
(i) The range of )
(x
f
(ii) Sketch the graph of )
(x
f
31. When a motor-cyclist is travelling along a straight stretch of road from
south to north at a steady speed of 30 km per hour, the wind appears to him
to come from a direction E
N 
40 . When he returns along the same road at
the same speed the wind appears to come from a direction E
S 
30 . Find,
by drawing or otherwise the true magnitude and direction of the wind.
32. Prove that ( )( )
1
2
sin
2
cos
sin
3
cos
3
sin −
+
=
− A
A
A
A
A , hence or
otherwise, find the value of A such that 


 180
0 A for which
( ) ( )
A
A
A
A cos
sin
2
3
cos
3
sin
3 +
=
− .
33. In the quadrilateral ABCD, 13
=
AB cm, 20
=
BC cm, 48
=
CD cm and
DBC
BAC 
=
 . Without using tables
(a) Show that
13
5
cos =
BAC
(b) Show that
5
4
cos =
ACB
(c) Show that AD
AB =
(d) Calculate the area of the quadrilateral ABCD.
34. (a) Express 
 sin
2
cos
3 + in the form ( )
A
R −

cos , where 0

R and



 90
0 A , state the value of R
(b) Solve the equation 5
.
3
sin
2
cos
3 =
+ 
 giving your answer in the
interval 


 180
0 
(c) Sketch the graph of 
 sin
2
cos
3 +
=
y for 


 180
0 
35. (a) From a point P in a horizontal plane a man observes the summit S of a
mountain to bear due north at an elevation  When the man has walked a
distance 2a on a bearing  east of north to a point Q in the horizontal

BARAKA LO1BANGUT1
plane he observes that the elevation of S from Q is again  . If h is the
height of S above the horizontal plane containing P and Q show that

sec
θ
tan
a
h =
(c) When the man has walked a further distance a in the same plane and in the
same direction to a point R he observes that the elevation of S from R is  .
Show that ( ) 

 2
2
2
cot
1
cos
3
cot +
= and that the distance RS is
( )
3
sec
sec 2
3
+


a
36. (a) Solve the equation 5
tan
sec
2 2
=
− x
x for 


 360
0 x
(b) Show that
x
x
x
x
cos
cos
1
tan
sin
2
−
= hence or otherwise solve
3
tan
sin
2 =
x
x for 


 360
0 x
37. Prove that ( )
A
A
A 4
cos
1
8
1
cos
sin 2
2
−
= hence find  

 d
2
2
cos
sin
38. Find all real values of x and y that satisfy the simultaneous equations





=
+
=
+
259
tan
27
sin
64
35
tan
9
sin
16
2
2
y
x
y
x
where x, y are the measures of the angles
expressed in degrees.
39. Prove that A
A
A 2
cosec
2
cot
cot =
−
40. Prove that 
4
1
8
1
tan
2
7
1
tan
5
1
tan
2 1
1
1
=






+






+





 −
−
−
41. (a) If ( ) ( ) 

 2
tan
2
45
tan
45
tan =
−

−
+
 (b) Prove that
2
2
2
cot
1
tan
1
cot
1
tan
1






−
−
=
+
+
x
x
x
x
42. (a) Prove that
4
π
239
1
tan
119
120
tan
239
1
tan
5
1
tan
4 1
1
1
1
=






−






=






−





 −
−
−
−
(b) Using the expansion ...
5
1
3
1
tan 5
3
1
+
+
−
=
−
x
x
x
x evaluate π to five
decimal places.

BARAKA LO1BANGUT1
43. Given that
2
3
=
x
cos and 


 360
0 x . Find without using calculator or
table the value of x
tan and x
sin
44. (a) Prove that
4
π
239
1
tan
119
120
tan
239
1
tan
5
1
tan
4 1
1
1
1
=






−






=






−





 −
−
−
−
(b) Using the expansion ...
5
1
3
1
tan 5
3
1
+
+
−
=
−
x
x
x
x evaluate π to five
decimal places.
45. Let 
sin
=
x or otherwise, show that i
x
x
x
x
−
=
−
−
+
−
+
+
2
2
2
2
1
1
1
1
46. (a) Solve for x if 0
1
2
2
4 =
+
+
+ x
x
x
x cos
sin
cos
sin
(b) Find the least difference between the roots, in the first quadrant of the
equation ( ) ( ) 0
1
2
3
2
4 2
=
+
+
− x
x
x cos
sin
cos
(c) Simplify








−
−
−
−
+
+
−
x
x
x
x
sin
sin
sin
sin
cot
1
1
1
1
1
(d) Given 
=
+
+ −
−
−
180
1
1
1
z
y
x sin
sin
sin show that
( ) ( )
2
2
2
2
2
2
2
4
4
4
2
4 x
z
z
y
y
x
xyz
z
y
x +
+
=
+
+
+
(e) By expressing A
A sin
cos 2
+ in the form of ( )
B
A
Q +
sin where B is an
acute angle, find the minimum and maximum values of the expression and
hence find the values of A and B.
(f) Show that ( )
B
A
C
A
C
B
C
B
A
C
B
A
C
B
A
tan
tan
tan
tan
tan
tan
tan
tan
tan
tan
tan
tan
tan
−
−
−
−
+
+
=
+
+
1
hence show that C
B
A
C
B
A tan
tan
tan
tan
tan
tan =
+
+ if A, B and C are
angles of the triangles.
47. (a) If ( )
4
3
arcsin
2 =
A , without using a calculator show that
2
1
cos
sin 
=
− A
A
(b) Prove the following identities
i) 



cot
cosec
cos
1
cos
1
+
=
−
+

BARAKA LO1BANGUT1
ii)
( )
A
A
A
A
A
cot
cos
1
sin
sin
cos
1 2
=
+
−
+
(c) Express 
 sin
3
cos
2 + in the form of ( )

 +
sin
R then find
i) Maximum and minimum values
ii) The value of  for which the maximum and minimum values
occur.
(d) If B
A
p cos
cos −
= and B
A
q sin
sin −
= , express ( )
B
A −
cos and
( )
B
A +
sin in terms of p and q.

BARAKA LO1BANGUT1
TRIGONOMETRY
7
BARAKA LO1BANGUT1
Email: barakaloibanguti@gmail.com
Tel: +255 621842525

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